Question

If tan alpha = -2 find the values of sin alpha (alpha lies in 2 quadrant)

Answer 1

$\textbf{Given:}$

$tan\;\alpha=-2$

$\text{Taking reciprocals, we get}$

$cot\;\alpha=\frac{1}{-2}$

$\text{using}$

$\boxed{\bf\;cosec^2A-cot^2A=1}$

$cosec^2\alpha=1+cot^2\alpha$

$cosec^2\alpha=1+\frac{1}{4}$

$cosec^2\alpha=\frac{5}{4}$

$\text{Taking reciprocals, we get}$

$sin^2\alpha=\frac{4}{5}$

$sin\;\alpha=\pm\frac{2}{\sqrt{5}}$

$\text{since \alpha lies in second quadrant, sin\;\alpha is positive }$

$\therefore\boxed{\bf\;sin\;\alpha=\frac{2}{\sqrt{5}}}$

Answer 2

Answer:

\textbf{Given:}Given:

tan\;\alpha=-2tanα=−2

\text{Taking reciprocals, we get}Taking reciprocals, we get

cot\;\alpha=\frac{1}{-2}cotα=−21

\text{using}using

\boxed{\bf\;cosec^2A-cot^2A=1}cosec2A−cot2A=1

cosec^2\alpha=1+cot^2\alphacosec2α=1+cot2α

cosec^2\alpha=1+\frac{1}{4}cosec2α=1+41

cosec^2\alpha=\frac{5}{4}cosec2α=45

\text{Taking reciprocals, we get}Taking reciprocals, we get

sin^2\alpha=\frac{4}{5}sin2α=54

sin\;\alpha=\pm\frac{2}{\sqrt{5}}sinα=±52

\text{since $\alpha$ lies in second quadrant, $sin\;\alpha$ is positive }since α lies in second quadrant, sinα is positive 

\therefore\boxed{\bf\;sin\;\alpha=\frac{2}{\sqrt{5}}}∴sinα=52